If $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $( x^{4}+x^{3}-23 x^{2}=3 x+60)$, find the all zeroes of given polynomial.

Academic Mathematics NCERT Class 10

Given: $\sqrt{3}$ and $-\sqrt{3}$ are the zeroes of $( x^{4}+x^{3}-23 x^{2}=3 x+60)$.

To do: To find the all zeroes of given polynomial.

Solution:

Let $f( x)=x^4+x^3-23x^2-3x+60$

As given $\sqrt{3}$ and $-\sqrt{3}$ are the zeros of the polynomial.

$( x-\sqrt{3})$ and $(x+\sqrt{3})$ are factors of $f( x)$.

$( x-\sqrt{3})(x+\sqrt{3})$ also will be the factor of $f( x)$.

On dividing $f( x)$ by $( x^2-3)$.

Let $f( x)=0$.

$( x^2+x-20)( x^2-3)=0$

$\Rightarrow ( x^2+5x-4x-20)( x^2-3)=0$

$\Rightarrow [ x( x+5)-4( x+5)]( x^2-3)=0$

$\Rightarrow ( x-4)( x+5)( x+3)( x-3)=0$

$\Rightarrow x=4$ or $x=-5$ or $x=\sqrt{3}$ or $x=-\sqrt{3}$

Hence, all the zeros of the given polynomial are $\sqrt{3},\ -\sqrt{3},\ 4$ and $-5$.

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Updated on: 10-Oct-2022

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